ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE

GROUPS

PETE L. CLARK

1. Reminder on quotient groups

Let G be a group and H a subgroup of G. We have seen that the left cosets xHof H in G give a partition of G. Motivated by the case of quotients of rings byideals, it is natural to consider the product operation on cosets. Recall that for anysubsets S, T of G, by ST we mean {st | s ∈ S, t ∈ T}.

If G is commutative, the product of two left cosets is another left coset:

(xH)(yH) = xyHH = xyH.

In fact, what we really used was that for all y ∈ G, yH = Hy. For an arbitrarygroup G, this is a property of the subgroup H, called normality. But it is clear –and will be good enough for us – that if G is commutative, all subgroups are normal.

If G is a group and H is a normal subgroup, then the set of left cosets, denotedG/H, itself forms a group under the above product operation, called the quotientgroup of G by H. The map which assigns x ∈ G to its coset xH ∈ G/H is in facta surjective group homomorphism q : G → G/H, called the quotient map (or incommon jargon, the “natural map”), and its kernel is precisely the subgroup H.

Theorem 1. (Isomorphism theorem) Let f : G → G′ be a surjective homomor-phism of groups, with kernel K. Then G/K is isomorphic to G′.

Proof. We define the isomorphism q(f) : G/K → G′ in terms of f : map the cosetxK to f(x) ∈ G′. This is well-defined, because if xK = x′K, then x′ = xk for somek ∈ K, and then

f(x′) = f(x)f(k) = f(x) · e = f(x),

since k is in the kernel of f . It is immediate to check that q(f) is a homomorphismof groups. Because f is surjective, for y ∈ G′ there exists x ∈ G such that f(x) = yand then q(f)(xK) = y, so q(f) is surjective. Finally, if q(f)(xK) = e, thenf(x) = e and x ∈ K, so xK = K is the identity element of G/K. �

In other words, a group G′ is (isomorphic to) a quotient of a group G iff thereexists a surjective group homomorphism from G to G′.

Corollary 2. If G and G′ are finite groups such that there exists a surjective grouphomomorphism f : G→ G′, then #G′ | #G.

Proof. G′ ∼= G/ ker f , so #G′ ·#(ker f) = #G. �1

2 PETE L. CLARK

Remark: Suitably interepreted, this remains true for infinite groups.

Corollary 3. (“transitivity of quotients”) If G′ is isomorphic to a quotient groupof G and G′′ is isomorphic to a quotient group of G′, then G′′ is isomorphic to aquotient group of G.

Proof. We have surjective group homomorphisms q1 : G → G′ and q2 : G′ → G′′,so the composition q2 ◦ q1 is a surjective group homomorphism from G to G′′. �

2. Cyclic groups

Recall that a group G is cyclic if there exists some element g in G such that everyx in g is of the form gn for some integer n. (Here we are using the conventions thatg0 = e is the identity element of G and that g−n = (g−1)n.) Such an element g iscalled a generator. In general, a cyclic group will have more than one generator,and it is a number-theoretic problem to determine how many generators there are.

Example 1: The integers Z under addition are a cyclic group, because 1 is a gener-ator. The only other generator is −1.

Example 2: We denote by Zn the additive group of the ring (Z/nZ). It is alsoa cyclic group, because it is generated by the class of 1 (mod n).

We claim that these are the only cyclic groups, up to isomorphism. One (com-paratively sophisticated) way to see this is as follows: let G be a cyclic group, withgenerator g. Then there is a unique homomorphism f from the additive group ofthe integers to G which maps 1 to g. The map f is surjective because, by assump-tion, every y in G is of the form gn for some n ∈ Z, i.e., y = gn = f(n). Let K bethe kernel of this homomorphism. Then it is a subgroup of (Z,+), and since everyadditive subgroup of (Z,+) is an ideal, we have K = nZ for some n ∈ N. Thereforeby the isomorphism theorem, we have that G is isomorphic to the additive groupof the quotient ring Z/nZ, i.e., to Zn.

Corollary 4. Every quotient group of a cyclic group is cyclic.

Proof. We saw that a group is cyclic iff it is isomorphic to a quotient of (Z,+).Therefore a quotient G′ of a cyclic group is a group that is isomorphic to a quotientof a quotient of (Z,+), and by Corollary 3 this simply means that G′ is isomorphicto a quotient of (Z,+) and hence is itself cyclic. �Proposition 5. Let n ∈ Z+. For every positive divisor k of n, there is a uniquesubgroup of Zn of order k, and these are the only subgroups of Zn.

Proof. For any divisor k of n, the subgroup generated by k (mod n) of (Z/nZ,+)has order n

k , and as k runs through the positive divisors of n so does nk . So there is

at least one cyclic subgroup of Zn of order any divisor of n. Conversely, let H bea subgroup of (Z/nZ,+) and let k be the least positive integer such that the classof k mod n lies in H. (Since the class of n lies in H, there is such a least integer.)I leave it to you to show that H is the subgroup generated by k (mod n). �Remark: Slicker is to observe that the subgroups of Zn correspond to the ideals inZ/nZ which – by a general principle on ideals in quotient rings – correspond to theideals of Z containing (nZ), which correspond to the positive divisors of n.

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 3

Corollary 6. Subgroups of cyclic groups are cyclic.

Proposition 7. For a ∈ Z+, the order of the class of a ∈ (Z/nZ,+) is ngcd(a,n) .

Proof. Let d = gcd(a, n) and write a = da′. The (additive) order of a (mod n) isthe least positive integer k such that n | ka. We have n | ka = kda′ ⇐⇒ n

d | ka′,and since gcd(nd , a) = 1, the least such k is n

d . �

Corollary 8. Let a ∈ Z, n ∈ Z+.a) The class of a ∈ Zn is a generator if and only if gcd(a, n) = 1. In particularthere are φ(n) generators.b) For any d | n, there are precisely φ(d) elements of Zn of order d.c) It follows that

∑d | n φ(d) = n.

Proof. Part a) is immediate from Proposition 7. For any d | n, each element of orderd generates a cyclic subgroup of order d, and we know that there is exactly onesuch subgroup of Zn, so the elements of order d are precisely the φ(d) generators ofthis cyclic group. Part c) follows: the left hand side gives the number of elementsof order d for each d | n and the right hand side is #Zn. �This leads to a very useful result:

Theorem 9. (Cyclicity criterion) Let G be a finite group, not assumed to be com-mutative. Suppose that for each n ∈ Z+, there are at most n elements x in G withxn = e. Then G is cyclic.

Proof. Suppose G has order N , and for all 1 ≤ d ≤ N , let f(d) be the numberof elements of G of order d. By Lagrange’s Theorem, f(d) = 0 unless d | N , soN = #G =

∑d | N f(d). Now, if f(d) ̸= 0 then there exists at least one element of

order d, which therefore generates a cyclic group of order d, whose elements gived solutions to the equation xd = e. By our assumption there cannot be any moresolutions to this equation, hence all the elements of order d are precisely the φ(d)generators of this cyclic group. In other words, for all d |n we have either f(d) = 0or f(d) = φ(d), so in any case we have

N =∑d | N

f(d) ≤∑d | N

φ(d) = N.

Therefore we must have f(d) = φ(d) for all d | N , including d = N , i.e., there existsan element of G whose order is the order of G: G is cyclic. �Corollary 10. Let F be a field, and let G ⊂ F× be a finite subgroup of the groupof nonzero elements of F under multiplication. Then G is cyclic.

Proof. Indeed, by basic field theory, for any d ∈ Z+ the degree d polynomial td − 1can have at most d solutions, so the hypotheses of Theorem 9 apply to G. �

3. Products of elements of finite order in a commutative group

Let G be a commutative group, and let x, y ∈ G be two elements of finite order, sayof orders m and n respectively. There is a unique smallest subgroup H = H(x, y) ofG containing both x and y, called the subgroup generated by x and y. H(x, y)is the set of all elements of the form xayb for a, b ∈ Z. Moreover, since x has orderm and y has order n, we may write every element of H as xayb with 0 ≤ a < m,0 ≤ b < n, so that #H ≤ mn. In particular the subgroup of an abelian group

4 PETE L. CLARK

generated by two elements of finite order is itself finite.

It is very useful to have some information about both the size of H(x, y) andthe order of the element xy in terms of m and n alone. However we cannot expecta complete answer:

Example 3: Suppose that m = n = N . We could take G to be the additivegroup of Z/NZ× Z/NZ, x = (1, 0), y = (0, 1). Then the subgroup generated by xand y is all of G, so has order N2, and the order of x+ y is N . On the other hand,we could take G = ZN and x = y = g some generator. Then H(x, y) = G has orderN and #xy is N if N is odd and N

2 is N is even. Or we could have taken y = x−1

so that the xy = e and has order 1. And there are yet other possibilities.

Example 4: Suppose that gcd(m,n) = 1. We can show that xy has order mn,and hence is a generator for H(x, y). Indeed, let a ∈ Z+ be such that (xy)a = e,i.e., xa = y−a. But the order of xa divides m and the order of y−a divides n; sincegcd(m,n) = 1, xa = y−a = 1, so that a | m, a | n. Since, again, gcd(m,n) = 1,this implies a | mn.

The general case is as follows:

Theorem 11. Let x and y be elements of finite order m and n in a commutativegroup G. Denote by H(x, y) the subgroup generated by x and y.a) lcm(m,n) | #H(x, y) | mn.b) lcm(m,n)

gcd(m,n) | #(xy) | lcm(m,n).

Proof. Step 1: We can define a surjective homomorphism of groups Ψ : Zm×Zn →H(x, y) by (c, d) 7→ xcy−d, so by #H(x, y) | #(Zm × Zn) by Corollary 2.

Step 2: Let K be the kernel of Ψ. By the Isomorphism theorem, #H(x, y) =#(Zm × Zn)/#K = mn

#K , so #H(x, y) | mn. Moreover, the kernel K consists of

pairs (c, d) such that xc = yd. Let f = gcd(m,n). Let o be the order of xc = yd.Since the order of xc divides m and the order of yd divides n, o | gcd(m,n) = f .There are f values of c (mod m) for which xc has order dividing f , and for each ofthese values, there is at most one value of d (mod n) such that xc = yd (becausethe elements yi for 0 ≤ i < n are distinct elements of G). This shows that thekernel can be viewed as a subset of Zf , and it is easily checked to be a subgroup.So #K | f and hence

lcm(m,n) =mn

f| mn#K

= #H(x, y).

Step 3: (xy)lcm(m,n) = xlcm(m,n)ylcm(m,n) = 1, so the order of xy divides lcm(m,n).

Step 4: Finally, suppose that a ∈ Z+ is such that (xy)a = xaya = 1, so xa = y−a.So the order of xa, which is m

gcd(a,m) is equal to the order of y−a, which is ngcd(a,n) .

In other words, we have

m gcd(a, n) = n gcd(a,m).

Since gcd(mf , n) = 1, mf | gcd(a,m), or

m | f gcd(a,m) | fa.

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 5

Similarly

n | f gcd(a, n) | fa.

Therefore lcm(m,n) | fa, or lcm(m,n)gcd(m,n) | a, completing the proof of the theorem. �

Remark: The divisibilities in Theorem 11 are best possible: if h and o are positive

integers such that lcm(m,n) | h | mn and lcm(m,n)gcd(m,n) | o | lcm(m,n), then there exist

elements x, y ∈ Zm × Zn such that #H(x, y) = h, #xy = o.

Remark: The situation is profoundly different for noncommutative groups: forevery m,n ≥ 2 and 2 ≤ r ≤ ∞ there exists a group G containing elements x oforder m, y of order n whose product xy has order r. Moreover, if r <∞ then onecan find a finite group G with these properties, whereas one can find an infinitegroup with these properties iff 1

m + 1n + 1

r ≤ 1.

The following is a consequence of Theorem 11 (but is much simpler to prove):

Corollary 12. Let m,n ∈ Z+. The group Zm × Zn is cyclic iff gcd(m,n) = 1.

Proof. The order of any element (c, d) divides lcm(m,n), and the order of (1, 1) islcm(m,n). Therefore the group is cyclic iff mn = lcm(m,n) iff gcd(m,n) = 1. �

4. Character theory of finite abelian groups

4.1. Introduction.

In this section our goal is to present the theory of characters of finite abeliangroups. Although this is an “easy” theory in that we can present it in its entiretyhere, it nevertheless of the highest impotance, being the jumping off point for atleast two entire disciplines of mathematics: the general theory of linear represen-tations of groups, and Fourier analysis. The special case of characters of the unitgroups U(N) = (Z/NZ)× will be used as one of the essential ingredients in theproof of Dirichlet’s theorem on primes in arithmetic progessions.

Let G be a finite commutative group. A character χ : G→ C× of G is a homomor-phism from G to the group C× of nonzero complex numbers under multiplication.

Suppose N = #G. By Lagrange’s theorem we have, for any g ∈ G, that gN = e(the identity element), and thus for any character χ on G we have

χ(g)N = χ(gN ) = χ(e) = 1.

Thus χ(g) is itself a complex Nth root of unity. Recall that the set of all complexNth roots of unity forms a cyclic group of order N , say µN . In other words, everycharacter on a group G of order N is really just a homomorphism from G to µN ,or equally well, from G into any fixed order N cyclic group.

We write X(G) for the set of all characters of G. We can endow X(G) with thestructure of a group: given χ1, χ2 ∈ X(G), we define their product “pointwise”:

∀g ∈ G, (χ1χ2)(g) := χ1(g)χ2(g).

6 PETE L. CLARK

The identity element is the trivial character g 7→ 1 for all g, and the inverse of χis the function χ−1 : g 7→ 1

χ(g) . Because for any z ∈ C we have zz = |z|2, if if z is a

root of unity, then the inverse of z is given by its complex conjugate z. It followsthat the inverse of a character χ is also given by taking complex conjugates:

χ(g) = χ(g) =1

χ(g)= χ−1(g).

4.2. The Character Extension Lemma.

Most of the content of the entire theory resides in the following result.

Lemma 13. (Character Extension Lemma) Let H be a subgroup of a finite com-mutative group G. For any character ψ : H → C×, there are [G : H] charactersΨ : G→ C× such that Ψ|H = ψ.

Proof. The result is clear if H = G, so we may assume that there exists g ∈ G \H.Let Gg = ⟨g,H⟩ be the subgroup generated by H and g. Now we may or may nothave Gg = G, but suppose that we can establish the result for the group Gg andits subgroup H. Then the general case follows by induction, since for any H ⊂ Gchoose g1, . . . , gn such that G = ⟨H, g1, . . . , gn⟩. Then we can define G0 = H andfor 1 ≤ i ≤ n, Gi = ⟨Gi−1, gi⟩. Applying the Lemma in turn to Gi−1 as a subgroupof Gi gives that in all the number of ways to extend the character ψ of H = G0 is

[G1 : G0][G2 : G1] · · · [Gn : Gn−1] = [G : G0] = [G : H].

So let us now prove that the number of ways to extend ψ from H to Gg = ⟨H, g⟩is [Gg : H]. For this, let d be the order of g in G, and consider G̃ := H × ⟨g⟩.The number of ways to extend a character ψ of H to a character of G̃ is equal to#⟨g⟩ = d: such a homomorphism is uniquely specified by the image of (1, g) inµd ⊂ C×, and all d such choices give rise to homomorphisms.

Moreover, there is a surjective homomorphism φ : H × ⟨g⟩ to Gg: we just take(h, gi) 7→ hg−i. The kernel of φ is the set of all pairs (h, gi) such that gi = h. Inother words it is precisely the intersection H ∩ ⟨g⟩, which has cardinality, say e,some divisor of d. It follows that

#Hg =#H × ⟨g⟩#H ∩ ⟨g⟩

=d

e·#H,

so

[Hg : H] =d

e.

But a homomorphism f : H × ⟨g⟩ → C× descends to a homomorhpism on thequotientHg iff it is trivial on the kernel of the quotient map, i.e., is trivial onH∩⟨g⟩.In other words, the extensions of ψ to a character of Hg correspond precisely to the

number of ways to map the order d element g into C× such that gde gets mapped

to 1. Thus we must map g to a de th root of unity, and conversely all such mappings

induce extensions of ψ. Thus the number of extensions is de = [Hg : H]. �

Corollary 14. For any finite commutative group G, X(G) is finite and

#X(G) = #G.

Proof. Apply Lemma 13 with H = 1. �

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 7

Corollary 15. For G a finite commmutative group and g ∈ G, TFAE:(i) For every χ ∈ X(G), χ(g) = 1.(ii) g is the identity element e of G.

Proof. Certainly (ii) =⇒ (i). Conversely, if g ̸= e, then H := ⟨g⟩ is a nontrivialcyclic group. By Corollary 14, there exists a nontrivial character ψ of H. Sinceg generates H, this implies ψ(g) ̸= 1. Now apply Lemma 13 to extend ψ to acharacter of G. �From these results one can deduce that the character group construction behavesnicely under homomorphisms: suppose f : G → H is a homomorphism of finitecommutative groups. Then we can define a map X(f) : X(H) → X(G) – notewell: in the opposite direction! – just by taking a character χ : H → C× andprecomposing it with f to get a character χ ◦ f : G→ C×.

Proposition 16. Let f : G→ H be a homomorphism of finite commutative groups.a) The above map X(f) : X(H) → X(G) is a group homomorphism.b) The homomorphism f is injective ⇐⇒ the homomorphism X(f) is surjective.c) The homomorphism f is surjective ⇐⇒ the homomorphism X(f) is injective.

Proof. Part a) is a straightforward verification which we leave to the reader.b) Assume first that f is injective. We may as well assume then that G is a

subgroup of H and f = ι is the inclusion map. Then the induced homomorphismX(ι) : X(H) → X(G) is nothing else than the map which restricts a character ofH to a character of the subgroup G; that this restriction map is surjective is animmediate consequence of Lemma 13. Inversely, assume that f is not injective, sothat there exists e ̸= g ∈ G such that f(g) = e ∈ H. By Corollary 15, there exists acharacter χ : G→ C× such that χ(g) ̸= 1. But then for any character ψ : H → C×,we have

(ψ ◦ f)(g) = ψ(e) = 1,

which shows that ψ ◦ f ̸= χ, i.e., χ is not in the image of X(f).c) By the Extension Lemma, the number of characters on H which are trivial

on f(G) is [H : f(G)]. Therefore this quantity is equal to 1 – i.e., f is surjective –iff a character ψ on H for which ψ ◦ f is trivial is necessarily itself trivial. �4.3. Orthogonality relations.

Theorem 17. Let G be a finite abelian group, with character group G.a) For any nontrivial character χ ∈ X(G), we have

∑g∈G χ(g) = 0.

b) For any nontrivial element g of G, we have∑

χ∈X(G) χ(g) = 0.

Proof. a) Put

(1) S =∑g∈G

χ(g).

Since χ is nontrivial, there exists g0 ∈ G such that χ(g0) ̸= 1. Multiplying bothsides of (1) by χ(g0), we get

χ(g0)S =∑g∈G

χ(g)χ(g0) =∑g∈G

χ(gg0) =∑g∈G

χ(g) = S;

the penultimate equality holds because, as g runs through all elements of G, sodoes g0. Therefore we have

(χ(g0)− 1)S = 0.

8 PETE L. CLARK

Since χ(g0) ̸= 1, the inexorable conclusion is that S = 0. As for part b), if g ̸= e,then by Corollary 15 there exists a character χ such that χ(g) ̸= 1, and then theargument is identical to part a).1 �

Let us briefly explain why these are called orthogonality relations. Consider the setCG of all functions f : G→ C. Under pointwise addition and scalar multiplication,CG is a complex vector space, of dimension equal to #G. We can define a Hermitianinner product on CG as follows:

⟨f, g⟩ := 1

#G

∑x∈G

f(x)g(x).

Now let χ1 and χ2 be characters of G. If χ1 = χ2, then we have

⟨χ1, χ1⟩ =1

#G

∑x∈G

|χ1(x)|2 = 1,

whereas if χ1 ̸= χ2, then χ1χ−12 is nontrivial, and then Theorem 17 gives

⟨χ1, χ2⟩ =1

#G

∑x∈G

(χ1χ−12 )(x) = 0.

In other words, the set X(G) of characters of G is orthonormal with respect to thegiven inner product. In particular, the subset X(G) of CG is linearly independent.Since its cardinality, #G, is equal to the dimension of CG, we conclude:

Corollary 18. Let G be a finite commutative group, and let CG be the C-vectorspace of all functions from G to C, endowed with the inner product

⟨f, g⟩ = 1

#G

∑x∈G

f(x)g(x).

Then the set of characters of G forms an orthonormal basis with respect to ⟨ , ⟩.Therefore, any function f : G→ C can be expressed as a unique linear combinationof characters. Explicitly:

f =∑

χ∈X(G)

⟨f, χ⟩χ.

This can be viewed as the simplest possible case of a Fourier inversion formula.

4.4. The canonical and illicit isomorphism theorems; Pontrjagin duality.

In the course of study of finite commutative groups, one sees that subgroups andquotient groups have many similar properties. For instance, subgroups of cyclicgroups are cyclic, and also quotients of cyclic groups are cyclic. Moreover, a cyclicgroup of order n has a unique subgroup of every order dividing n and no othersubgroups, and the same is true for its quotients. If one plays around for a bit withfinite commutative groups, one eventually suspects the following result:

Theorem 19. Let G and H be finite commutative groups. Then TFAE:(i) H can be realized as a subgroup of G: ∃ an injective homomorphism H → G.(ii) H can be realized as a quotient of G: ∃ a surjective homomorphism G→ H.

1Alternately, using the canonical isomorphism G ∼= X(X(G)) described in the next section,one can literally deduce part b) from part a).

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 9

There is a certain resemblance between Theorem 19 and Proposition 16, but theyare not the same. Proposition 16 asserts that if there is an injection H → G, thereis a surjection X(G) → X(H) (and similarly with “injection” and “surjection”interchanged). To deduce Theorem 19 from Proposition 16, one needs the following:

Theorem 20. (Illicit Isomorphism Theorem) Any finite commutative group G isisomorphic to its chracter group X(G).

Some cases of Theorem 20 are easy to establish. For instance, since G and X(G)have the same order, they must be isomorphic whenever #G is prime. Further, togive a character on a cyclic group of order N it suffices to send a fixed generatorto any Nth root of unity in C. More precisely, choosing a generator of an abstractcyclic group G order N amounts to choosing an isomorphism of G with Z/NZ (wesend the generator to 1 (mod N)). And the characters on Z/NZ are all obtainedby exponentiation: for any c ∈ Z/NZ, there is a unique character χa such that

χc(1) = e2πic/N

and therefore for any b ∈ Z/NZ

χc(b) = e2πicb/N .

It is immediate to check that χc ·χc′ = χc+c′ , where addition is taken mod N . Thus

we get a canonical isomorphism X(Z/NZ) ∼→ Z/NZ.

Moreover, if G1 and G2 are finite commutative groups, then in a natural way

X(G1 ×G2) = X(G1)×X(G2);

again we leave the details to the interested reader. Of course the analogous identityfor products of any finite number of groups follows by induction.

Combining these observations, it follows that G ∼= X(G) for any finite commu-tative group G of the form Zn1 × . . . × Znk

, i.e., for any direct product of cyclicgroups. Is this enough to prove Theorem 20? Indeed it is, because of the following:

Theorem 21. (Fundamental theorem on finite commutative groups) Let G be afinite commutative group.a) There exist prime powers pa1

1 , . . . , parr (we allow pi = pj for i ̸= j) such that

G ∼= Zpa11

× . . .× Zparr,

i.e., G is a direct product of finite cyclic groups of prime power order.b) Moreover, this decomposition is essentially unique in the following (familiar)sense: if also we have

G ∼= Zqb11

× . . .× Zqbss,

then r = s and there exists a bijection σ : {1, . . . , r} → {1 . . . s} such that for all1 ≤ i ≤ r, qσ(i) = pi and bσ(i) = ai.

Now please bear with me while I make a few possibly confusing remarks about whyI have labelled Theorem 20 the “illicit” isomorphism theorem. In some sense it is“lucky” that G ∼= X(G), in that it is not part of the general meaning of “duality”that an object be isomorphic to its dual object. Rather, what one has in muchmore generality is a canonical injection from an object to its double dual. Here,this means the following: we can construct a canonical map G → X(X(G)). In

10 PETE L. CLARK

other words, given an element g in G, we want to define a character, say g•, on thecharacter group, i.e., a homomorphism X(G) → C×. This may sound complicatedat first, but in fact there is a very easy way to do this: define g •χ := χ(g)! It is noproblem to check that the association g 7→ g• is a homomorphism of finite abeliangroups. Moreover, suppose that for any fixed g ∈ G the map g• were trivial: thatmeans that for all χ ∈ X(G), χ(g) = 1. Applying Corollary 15, we get that g = 1.Therefore this map

• : G→ X(X(G))

is an injective homomorphism between finite abelian groups. Moreover,

#X(X(G)) = #X(G) = #G,

so it is an injective homomorphism between finite groups of the same order, andtherefore it must be an isomorphism.

In order to write down the isomorphism •, we did not have to make any choices.There is a precise sense in which the isomorphism to the double dual is “canonical”and any isomorphism between G and X(G) is “noncanonical”, but explanining thisinvolves the use of category theory so is not appropriate here. More interesting is toremark that there is a vastly more general class of commutative groups G for whichX(G) is defined in such a way as to render true all of the results we have proved hereexcept the illicit isomorphism theorem: we need not have G ∼= X(G). For this wetake G to be a commutative group endowed with a topology which makes it locallycompact Hausdorff. Any commutative group G can be endowed with the discretetopology, which gives many examples. For a finite group the discrete topology is theonly Hausdorff topology, so this is certainly the right choice, but an infinite groupmay or may not carry other interesting locally compact topologies. Some examples:

Example 1: The integers Z: here we do want the discrete topology.

Example 2: The additive group (R,+) with its usual Euclidean topology: thisis a locally compact group which is neither discrete nor compact. More gener-ally, one can take (Rn,+) (and in fact, if G1 and G2 are any two locally compactcommutative groups, then so is G1×G2 when endowed with the product topology).

Example 3: The multiplicative group C× of the complex numbers is again lo-cally compact but neither discrete nor compact, but it is “closer to being compact”then the additive group C ∼= R2. In fact, considering polar coordinates gives anisomorphism of topological groups C× ∼= R>0 × S1, where S1 is the unit circle.Moreover, the logarithm function shows that R>0 is isomorphic as a topologicalgroup to (R,+), so all in all C× ∼= (R,+) × S1. Note that S1, the circle group, isitself a very interesting example.

Now, given any locally compact commutative group G, one defines the Pontrjagindual group X(G), which is the group of all continuous group homomorphismsfrom G to the circle group S1. Moreover, X(G) can be endowed with a natural

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 11

topology.2 Again, one has a natural map G→ X(X(G)) which turns out to be anisomorphism in all cases.

If G is a finite, discrete commutative group, then as we saw, any homomorphismto C× lands in S1 (and indeed, the countable subgroup of S1 consisting of all rootsof unity) anyway; moreover, by discreteness every homomorphism is continuous.Thus X(G) in this new sense agrees with the character group we have defined. Butfor infinite groups Pontrjagin duality is much more interesting: it turns out thatG is compact iff X(G) is discrete.3 Since a topological space is both compact anddiscrete iff it is finite, we conclude that a topological group G which is infinite andeither discrete or compact cannot be isomorphic to its Pontrjagin dual.

In our examples, it is easy to see that Hom(Z, S1) = S1, which according to thegeneral theory implies Hom(S1, S1) = Z: that is, the discrete group Z and the com-pact circle group S1 are mutually dual groups. This is the theoretical underpinningof Fourier series.

However, if G is neither discrete nor compact, then the same holds for X(G),so there is at least a fighting chance for G to be isomorphic to X(G). Indeed thishappens for R: Hom(R, S1) = R, where we send x ∈ R to the character t 7→ e2πitx.4

This is the theoretical underpinning of the Fourier tranform.

Another sense in which the isomorphism between G and X(G) for a finite com-mutative group G is “illicit” is that turns out not to be necessary in the standardnumber-theoretic applications. A perusal of elementary number theory texts re-veals that careful authors take it as a sort of badge of honor to avoid using theillicit isomorphism, even if it makes the proofs a bit longer. For example, the mostnatural analysis of the group structure of (Z/2aZ)× for a ≥ 3 would consist inshowing: (i) the group has order 2a−1; (ii) it has a cyclic subgroup of order 2a−2;(iii) it has a noncyclic quotient so is itself not cyclic. Applying Theorem 21 onecan immediately conclude that it must be isomorphic to Z2a−2 × Z2. In our workin Handout 9.5, however, we show the isomorphism by direct means.

This was first drawn to my attention by a close reading of J.-P. Serre’s text[Se73]5 in which the illicit isomorphism is never used. Indeed, the careful readerwill see that, following Serre, our main application of character groups – namelythe proof of Dirichlet’s theorem on primes in arihtmetic progressions – uses only#X(G) = #G, but not X(G) ∼= G.

2If you happen to know something about topologies on spaces of functions, then you know

that there is one particular topology that always has nice properties, namely the compact-opentopology. That is indeed the correct topology here.

3Similarly, G is discrete iff X(G) is compact; this follows from the previous statement togetherwith G ∼= X(X(G)).

4Similarly Rn is self-dual for any n.5This is a wonderful book, but don’t be fooled by the name: it is a graduate level text in

number theory!

12 PETE L. CLARK

However, to my mind, avoiding the proof of Theorem 21 gives a misleading im-pression of the difficulty of the result.6 On the other hand, Theorem 21 evidentlyhas some commonalities with the fundamental theorem of arithmetic, which makesit somewhat desirable to see the proof. In the next section we provide such a proof,which is not in any sense required reading.

5. Proof of the Fundamental theorem on finite commutative groups

First some terminology: Let G be a commutative group, written multiplicatively.

If #G = pa is a prime power, we say G is a p-group.

For n ∈ Z+, we put G[n] = {x ∈ G | xn = 1}. This is a subgroup of G.

We say that two H1, H2 subgroups of G are complementary if H1 ∩H2 = {1},H1H2 = G. In other words, every element g of G can be uniquely expressed inthe form h1h2, with hi ∈ Hi. In yet other (equivalent) words, this means preciselythat the homomorphism H1 × H2 → G, (h1, h2) 7→ h1h2 is an isomorphism. Wesay that a subgroup H is a direct factor of G if there exists H ′ such that H,H ′

are complementary subgroups. Thus, in order to prove part a) it suffices to showthat every finite commutative group has a nontrivial direct factor which is cyclic ofprime power order; and in order to prove part b) it suffices (but is much harder!)to show that if G ∼= H ×H ′ ∼= H ×H ′′ then H ′ ∼= H ′′.

More generally if we have a finite set {H1, . . . , Hr} of subgroups of G such thatHi ∩Hj = {1} for all i ̸= j and G = H1 · · ·Hr, we say that the Hi’s form a set ofcomplementary subgroups and that each Hi is a direct factor. In such a circum-stance we have G ∼= H1 × . . .×Hr.

We now begin the proof of Theorem 21.

Step 1 (primary decomposition): For any commutative group G, let Gp be the

set of elements of G whose order is a power of p. Also let Gp′be the set of elements

of G whose order is prime to p. It follows from Theorem 11b) that Gp and Gp′are

both subgroups of G. We claim that Gp and Gp′are complementary subgroups.

Certainly Gp ∩ Gp′= {e}, since any element of the intersection would have both

order a power of p and relatively prime to p and thus have order 1 and be theidentity. On the other hand, let x be any element of G, and write its order as pk · bwith gcd(p, b) = 1. Thus we can choose i and j such that ipk + jb = 1, and then

x = x1 = xipk+jb = (xp

k

)i · (xb)j , and by Proposition 7 the order of (xpk

)i dividesb (so is prime to p) and the order of (xb)j divides pk. This proves the claim. Nowa simple induction argument gives the following:

Proposition 22. Let G be a finite abelian group, of order n = pa11 · · · par

r . Thenthe subgroups {Gpi}ri=1 form a set of complementary subgroups, and the canonicalmap H1 × . . .×Hr → G, (h1, . . . , hr) 7→ h1 · · ·hr is an isomorphism of groups.

6The real reason it is often omitted in such treatments is that the authors know that they

will be giving a more general treatment of the structure theory finitely generated modules over aprincipal ideal domain, of which the theory of finite commutative groups is a very special case.

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 13

Thus any finite commutative group can be decomposed, in a unique way, into a di-rect product of finite commutative groups of prime power order. We may thereforeassume that G is a commutative p-group from now on.

Step 2: We prove the following refinement of Theorem 9 for commutative p-groups:

Proposition 23. Let p be a prime and G be a finite commutative group of orderpa for some a ∈ Z+. TFAE:(i) G has exactly p elements of order p.(ii) G is cyclic.

Proof. We already know that (ii) =⇒ (i), of course. Assume (i); the naturalstrategy is to appeal to our cyclicity criterion Theorem 9. In this case we wish toshow that for any 0 < k ≤ a, there are at most pk elements of G of order dividingpk. We accomplish this by induction (!); the case of k = 1 is our hypothesis, soassume that for all 1 ≤ ℓ < k the number of elements of order dividing pℓ in G isat most pℓ and we wish to show that the number of element of order dividing pk isat most pk. For this, consider the endomorphism

φ : G[pk] → G[pk], x 7→ xpk−1

.

Now the kernel of φ is precisely G[pk−1], which we have inductively assumed hasorder at most pk−1. If the order of G[pk] exceeds pk, then since

φ(G[pk]) ∼= G[pk]/Ker(φ),

we would have #φ(G[pk]) > p. But by Proposition 7 the image of φ consists entirelyof elements of order dividing p, contradiction. �

Step 3:

Proposition 24. Let G be a finite commutative p-group, and let pa be the maximumorder of an element of G. Then every cyclic subgroup C of order pa is a direct factorof G: there exists a complementary subgroup H, giving an isomorphism G ∼= C×H.

Proof. The result holds vacuously for commutative groups of order p. Assume thatit holds for all commutative groups of order pk for k < a, and suppose we haveG = pa, x an element of maximal order in G and C = ⟨x⟩ If the order of x ispa, then G = C is cyclic and the conclusion again holds trivially. Otherwise, byProposition 23, there exists an order p subgroup K of G not contained in C, soC ∩K = {e}. Then the cyclic subgroup (C +K)/K has maximal order in G/K;by induction there exists a complementary subgroup H of G/K, i.e., a subgroupH containing K such that (C + K) ∩ H = K, (C + K) · H = G. It follows thatH∩C ⊂ K∩C = {e} and C ·H = G, so C andH are complementary subgroups. �

We may now deduce Theorem 21a) from Proposition 24. Indeed, given any finitep-group G we choose an element x of maximum order pa, which generates a cyclicsubgroup C of maximum order, which according to Proposition 24 has a com-plementary subgroup H and thus G ∼= Zpr ∼= H. Applying the same procedure toH, eventually we will express G as a product of finite cyclic groups of p-power order.

14 PETE L. CLARK

Step 4: Finally we address the uniqueness of the decomposition of a commuta-tive p-group into a direct product of cyclic groups.7 Suppose we have

G ∼= Zpa1 × . . .× Zpar∼= Zpb1 × . . .× Zpbs .

We may assume that a1 ≥ . . . ≥ ar and b1 ≥ . . . ≥ bs, and we wish to provethat r = s and ai = bi for all i. We may also inductively assume the uniquenessstatement for commutative p-groups of smaller order than G. Now let φ : G → Gbe x 7→ xp. Then we have

φ(G) ∼= Zpa1−1 × . . .× Zpar−1∼= Zpb1−1 × . . .× Zpbs−1 .

Since #φ(G) < #G, by induction the two decompositions are unique, the onlyproviso being that if an exponent ci is equal to 1, then Zpci−1 is the trivial group,which we do not allow in a direct factor decomposition. Therefore suppose that kis such that ai = 1 for all i > k and l is such that bj = 1 for all j > l. Then we getk = l and ai = bi for all 1 ≤ i ≤ k. But now we have

pr−k =#G

pa1+...+ak=

#G

pb1+...+bk= ps−k,

so we conclude r = s and thus ai = bi for 1 ≤ i ≤ r.

It is interesting to ask which of the steps go through for a group which is infi-nite, non-commutative or both.

Step 1 fails in a non-commutative group: the elements of p-power order need notform a subgroup. For instance, the symmetric group Sn is generated by transposi-tions. In any commutative group one can define the subgroups Gp for primes p, andthey are always pairwise disjoint. The subgroup they generate is called the torsionsubgroup of G and often denoted G[tors]: it consists of all elements of finite order.

Step 2 fails for noncommutative finite p-groups: The quaternion group Q8 ={±1,±i,±j,±k} is a noncyclic group of order 8 = 23 = p3 which has exactlyp = 2 elements of order dividing p. It is false for all infinite abelian groups, sincean infinite group can only be cyclic if its torsion subgroup is trivial.

Step 3 fails for finite noncommutative groups: again Q8 is a counterexample.

As for Step 4, one may ask the following

Question 1. Suppose we have three groups H, G1, G2 such that H×G1∼= H×G2.

Must it then be the case that G1∼= G2?

Without any restrictions the answer to this question is negative. For instance, onecan take H = G1 = (R,+), G2 = 0, and note that R × R ∼= R as Q-vector spaces,hence as commutative groups. On the other hand:

Theorem 25. (Remak-Krull-Schmidt) If H, G1 and G2 are all finite groups, thenindeed H ×G1

∼= H ×G2 implies G1∼= G2.

A group G is indecomposable if it is not isomorphic to H1 ×H2 with H1 and H2

both nonzero. By Theorem 21, a finite commutative group is indecomposable iff

7This part of the proof follows [Su95].

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 15

it is cyclic of prime power order. Any finite group can be written as a product ofindecomposable groups. Using Theorem 25 it can be shown that if

G ∼= H1 × . . .×Hr = K1 × . . .×Ks,

where each Hi andKj are indecomposable (nontrivial) groups, then r = s and thereexists a bijection σ : {1, . . . , r} → {1, . . . , r} such that Ki

∼= Hσ(i) for 1 ≤ i ≤ r.

6. Wilson’s Theorem in a Finite Commutative Group

Here is one of the classic theorems of elementary number theory.

Theorem 26. (Wilson’s Theorem) For an odd prime p, (p− 1)! ≡ −1 (mod p).

Remark: The converse of Wilson’s Theorem also holds: if for some integer n > 1we have (n− 1)! ≡ 1 (mod n), then n is prime. In fact it can be shown that for allcomposite n > 4, n | (n− 1)! (exercise).

Most of the standard proofs involve starting with an elementary group-theoreticfact and then recasting it to avoid group-theoretic language to a greater or lesserextent. Since this handout is meant to be a “comprehensive” guide to finite com-mutative groups, we may as well give the argument in its proper language.

For a finite group G, let d2(G) be the number of order 2 elements in G.

Theorem 27. (Wilson’s Theorem in a Finite Commutative Group)Let (G,+) be a finite commutative group, and let S =

∑x∈G x. Then:

a) If d2(G) ̸= 1, then S = 0.b) If d2(G) = 1 – so that G has a unique element, say t, of order 2 – then d2(G) = t.

Proof. We setG[2] = {x ∈ G | 2x = 0}.

Every nonzero element of G[2] has order 2, so by Theorem 21, we must have G[2] ∼=Z2 × . . .× Z2 = Zk

2 is a direct product of copies of the cyclic group of order 2.8

Consider the involution ι : G→ G given by x 7→ −x. The fixed points of ι – i.e.,the elements x ∈ G such that ι(x) = x – are precisely the elements of G[2]. Thus theelements of G \G[2] occur in pairs of distinct elements x,−x, so

∑x∈G\G[2] x = 0.

In other words,∑

x∈G x =∑

x∈G[2] x, and we are reduced to the case G[2] ∼= Zk2 .

Case 1: k = 0, i.e., G[2] = 0. Then∑x∈G[2]

x =∑

x∈{0}

x = 0.

Moreover, in this case d2(G) = 0, in agreement with the statement of the theorem.Case 2: k = 1, i.e., G[2] = Z2. Then∑

x∈G[2]

x =∑x∈Z2

x = 0 + 1 = 1,

where 1 is the unique element of order 2 in Z2∼= G[2] (and thus also the unique

element of order 2 in G). Again, this agrees with the statement of the theorem.

8Actually invocation of Theorem 21 is overkill here: any 2-torsion commutative group admitsthe unique structure of a vector space over the field F2 with 2 elements. Being finite, G[2] is

certainly finite-dimensional over F2, so is isomorphic as a vector space – hence a fortiori as anadditive group – to Fn

2 .

16 PETE L. CLARK

Case 3: k ≥ 2. Then d2(G) ≥ 3, so we wish to show S =∑

x∈Zk2x = 0. For each

1 ≤ i ≤ k, half of the elements of Zk2 have ith coordinate 0 ∈ Z2; the other half

have ith coordinate 1 ∈ Z2. So the sum of the ith coordinates of the elements of Zk2

is 2k/2 = 2k−1 = 0 ∈ Z2, since k ≥ 2: every coordinate of S equals 0, so S = 0. �We now show that Theorem 27 implies Theorem 26. Let F be a finite field. We takeG = F×, the multiplicative group of nonzero elements of F.9 Now x ∈ G[2] ⇐⇒x2 = 1, and the polynomial t2−1 has exactly two roots in any field of characteristicdifferent from 2 and exactly one root in any field of characteristic 2. So d2(F×) isequal to 1 if #F is odd and equal to 0 if #F is even. Thus:

Corollary 28. Let F be a finite field, put P =∏

x∈F× x. Then:a) If #F is even, then P = 1.b) If #F is odd, then P is the unique element of order 2 in F×, namely −1.So for any odd prime p, the second case holds for the field Z/pZ: Wilson’s Theorem.

As we mentioned above, Wilson’s Theorem construed as a statement about theproduct of all residue classes from 1 up to n− 1 modulo n holds exactly when n isprime. On the other hand, for composite n we may still apply Theorem 27 to thefinite commutative group U(n) = (Z/nZ)×.

Theorem 29. (Gauss) Let n > 2 be an integer, and let U(n) be the multiplicativegroup of units of the finite ring Z/nZ. Put P =

∏x∈U(n) x. Then:

a) We always have P = ±1 (mod n).b) In fact P = −1 (mod n) if and only if n is 4, an odd prime power pa, or twicean odd prime power 2pa.

Let us prove part a). Applying Theorem 27 to G = U(n), we see that P = 1(mod n) unless d2(G) = 1, in which case it is the unique element of order 2 in G.But for all n > 2 there is certainly at least one element of order 2 in U(n), namely−1 (mod n). So if it happens that there is exactly one such element, it must be−1, and thus we must have P = −1 (mod n).

To prove part b) we must determine for which n we have d2(U(n)) = 1. Since#U(n) = φ(n) is even for all n ≥ 2; thus, since U(n) \ U(n)[2] has even order, wemust have d2(U(n)) ≥ 1. Note further that d2(U(n)) = 1 when U(n) is cyclic, i.e.,when there exists a primitive root modulo n. Elsewhere in these notes10 we showthat primitive roots modulo n exist precisely when n is 4, an odd prime power ortwice an odd prime power: this proves half of Theorem 29b). For the other valuesof n U(n) is not cyclic, but there are noncyclic groups G with d2(G) = 1. So onehas to look a bit more carefully at the structure of the groups U(n) for general n.We leave the details to the interested reader.

Remark: After searching the literature for sources the material of this section,I found a paper of the early American group theorist George Abram Miller [Mi03].The parallel between this paper and the material of the present section is nearlyexact. In particular Miller proves Theorem 27 (his proof is remarkably close to theone given here) and applies it to prove Theorem 29. That this result was first stated

9Note well that we are now talking about multiplicative groups rather than additive groups.It makes no mathematical difference, of course, but the reader may wish to pause to reorient to

the new notation.10As of this writing, this takes place in a handout called A Word on Primitive Roots.

ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS 17

and proved by Gauss is not mentioned in Miller’s paper, but its title suggests thathe may have been aware of this.

References

[Mi03] G.A. Miller, A new proof of the generalized Wilson’s theorem. Ann. of Math. (2) 4 (1903),188–190.

[Se73] J.-P. Serre, A course in arithmetic. Translated from the French. Graduate Texts in Math-ematics, No. 7. Springer-Verlag, New York-Heidelberg, 1973.

[Su95] D.B. Surowski, The Uniqueness Aspect of the Fundamental Theorem of Finite AbelianGroups. Amer. Math. Monthly, 102 (1995), 162–163.